# Equation of line thru (-1, -4), perpendicular to 3x - 2y + 9 = 0

#### mech649

##### New member
Find the equation of the line that contains the point (-1, -4) and is perpendicular to the line 3x - 2y + 9 = 0

#### stapel

##### Super Moderator
Staff member
Find the equation of the line that contains the point (-1, -4) and is perpendicular to the line 3x - 2y + 9 = 0
What slope (here) did you obtain for the original line? For instance, did you solve for "y=" (here) and then read off the slope "m" from the form "y = mx + b"?

What slope, then, do you need to use for the perpendicular line?

Given this slope and the given points, what line equation (here) did you get?

#### pka

##### Elite Member
Find the equation of the line that contains the point (-1, -4) and is perpendicular to the line 3x - 2y + 9 = 0
Here is a useful result:
If $$\displaystyle Ax+By+C=0$$ is a line then the line $$\displaystyle Bx-Ay+(Ay_0-Bx_0)=0$$ is a line perpendicular to it through the point $$\displaystyle (x_0,y_0)$$.

Thank you!

#### mech649

##### New member
here is my work. My answer isnt the same as the answer in the book.

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#### stapel

##### Super Moderator
Staff member
here is my work.

$$\displaystyle 3x\,−\,2y\,+\,9\,=\,0$$

$$\displaystyle \dfrac{−2y}{−2}\,=\, \dfrac{−3x}{−2}\,−\,\dfrac{9}{−2}$$

$$\displaystyle y\,=\,\dfrac{−3x}{−2}\,\dfrac{−9}{−2}$$
I think you mean the following:

. . . . .$$\displaystyle \,3x\,−\,3x\,−\,2y\,+\,9\,−\, 9\, =\,0\, −\, 3x\, −\, 9$$

. . . . .$$\displaystyle −2y\, =\, −3x\, −\, 9$$

. . . . .$$\displaystyle −2y\, =\, −3x\, +\, (−9)$$

. . . . .$$\displaystyle \dfrac{−2y}{−2}\, =\, \dfrac{−3x}{−2}\, +\, \dfrac{−9}{−2}$$

But why not make things easier for yourself in the first place?

. . . . .$$\displaystyle 3x\, −\, 2y\, +\, 2y\, +\, 9\, =\, 0\, +\, 2y$$

. . . . .$$\displaystyle 3x\, +\, 9\, =\, 2y$$

Then do the division, but by a positive value, so it's harder to lose signs (as you did in the last quoted line above):

. . . . .$$\displaystyle \dfrac{3x}{2}\, +\, \dfrac{9}{2}\, =\, \dfrac{2y}{2}$$

. . . . .$$\displaystyle \dfrac{3}{2}\, x\, +\, \dfrac{9}{2}\, =\, y$$

$$\displaystyle y\,=\,\dfrac{−3x}{−2}\,\dfrac{−9}{−2}$$

$$\displaystyle y\, =\, \dfrac{2}{3}\, x\, +\, b$$
How on earth did you get the second line above from the first? Why did the slope change from 3/2 to 2/3? Where did the y-intercept go?

. . . . .$$\displaystyle y\, =\, \dfrac{3}{2}\, x\, +\, \dfrac{9}{2}$$